Question

Find the equation of the line that is perpendicular to 3x + 2y − 8 = 0 and passes through the mid-point of the line segment joining the points (5, −2) and (2, 2).

Answer 1

⇒ Rewriting the given equation 3x + 2y − 8 = 0 into the slope-intercept form, y = mx + c to find the slope (m₁):

2y = −3x + 8

`y = - 3/2 x + 4`

∴ The slope (m₁) of the given line is `- 3/2`.

⇒ Since the line is perpendicular, its slope (m₂) is the negative reciprocal:

`m_2 = - 1/m_1`

`m_2 = 2/3`

⇒ The line passes through the midpoint of the points (5, −2) and (2, 2), using the mid-point formula:

`M = ((x_1 + x_2)/2, (y_1 + y_2)/2)`

`M = ((5 + 2)/2, (-2 + 2)/2)`

`M = (7/2, 0)`

⇒ Using the point-slope formula with `m = 2/3  "and point" (7/2, 0)`:

y − y₁ = m(x − x₁)

`y - 0 = 2/3 (x - 7/2)`

`y = 2/3 x - 14/6`

`y = 2/3 x - 7/3`   ...[Multiply the entire equation by 3]

3y = 2x − 7

⇒ Rearranging the above equation in the standard form (Ax + By + C = 0),

2x − 3y − 7 = 0

Hence, the equation of the perpendicular line is 2x − 3y − 7 = 0.